Check sibling questions


Transcript

Question 3 Find the approximate value of f (5.001), where f (x) = x3 โ€“ 7x2 + 15.Let x = 5 and โˆ† x = 0.001 Given f (x) = x3 โ€“ 7x2 + 15 ๐‘“โ€™(x) = 3x2 โˆ’ 14x Now, โˆ†y = fโ€™(x) โˆ†๐‘ฅ = (3x2 โˆ’ 14x) 0.001 Also, โˆ†y = f (x + โˆ†x) โˆ’ f(x) f(x + โˆ†x) = f(x) + โˆ†y f (5.001) = x3 โˆ’ 7x2 + 15 + (3x2 โˆ’ 14x) 0.001 Putting value of x = 5 f (5.001) = 53 โˆ’ 7(5)2 + 15 + (0.001) [3"(5)2" โˆ’14(5)] = (125 โˆ’ 175 + 15) + (0.001) (5) = โˆ’35 + 0.005 = โˆ’34.995 Hence, the approximate value of f (5.001) is โˆ’34.995

  1. Chapter 6 Class 12 Application of Derivatives
  2. Serial order wise

About the Author

Davneet Singh

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 14 years. He provides courses for Maths, Science and Computer Science at Teachoo